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Math Help - Points on a graph (tangent line finding)

  1. #1
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    Points on a graph (tangent line finding)

    How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

    Given: y=x^2+2x

    Any help is welcomed
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  2. #2
    Junior Member ursa's Avatar
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    Quote Originally Posted by hemi View Post
    How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

    Given: y=x^2+2x

    Any help is welcomed
    hi hemi
    differentiate your function
    you will get dy/dx. it is the slope of the tangent.
    since in the question it has asked for horizontal line,means tangent parellel to x axis.
    therefore, its slope will be zero.
    put dy/dx=0
    and get the value of x
    put this value in the function and get the value of y.

    thank you
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  3. #3
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    Quote Originally Posted by ursa View Post
    hi hemi
    differentiate your function
    you will get dy/dx. it is the slope of the tangent.
    since in the question it has asked for horizontal line,means tangent parellel to x axis.
    therefore, its slope will be zero.
    put dy/dx=0
    and get the value of x
    put this value in the function and get the value of y.

    thank you
    Hi,

    Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best )

    Thanks!
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  4. #4
    Junior Member ursa's Avatar
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    Quote Originally Posted by hemi View Post
    Hi,

    Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best )

    Thanks!
    hi hemi
    this is your solution
    y=2*x+x^2
    dy/dx=2+2*x. this is the slope of the tangent.
    since tangent is parellel to x-axis therefore
    dy/dx=0
    => 2+2*x=0
    => x=-1
    put this value in function to get the value of y
    => y=-1
    therefore, the point on the function at which tangent of the function will be horizontal is (-1,-1).

    Thank You
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  5. #5
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    You could have done this, by the way, without using Calculus. This is a quadratic function so its graph is a vertical parabola. A tangent line will be horizontal only at its vertex. And we can get the vertex by completing the square: y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+1)^2- 1. For x= -1 we have y= 0^2+ 1= 1. For any other value of x, x+ 1 is NOT 0 so (x+ 1)^2 is positive and y is greater than 1. (-1, 1) is the vertex.
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